Have you tried proving that? That is, by the definitions and step by step.
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Asaf KaragilaJun 23 '11 at 18:18

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Implicit in your question is that the other operations are commutative and all are associative. This is true. Be careful that one-sided ideals can act a bit strangely.
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Jack SchmidtJun 23 '11 at 18:40

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@Asaf: probably the OP did try to prove it, but because it's false I imagine he had some trouble...
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Pete L. ClarkJun 23 '11 at 21:27

@Pete: When I get stuck and go seek the advice of others, be it online or in person, I usually mention what I had tried (unless I have a very good reason not to do so).
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Asaf KaragilaJun 23 '11 at 21:31

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@Asaf: The OP said "I see no reasons why the product of ideals should be commutative", which to me at least implies that he looked for a proof. Also to me, your comment sounds like it might be suggesting "It's very straightforward to prove, just follow your nose". Since in fact the statement is false, this seems a bit misleading.
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Pete L. ClarkJun 23 '11 at 21:35

2 Answers
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It is not true that ideal product is commutative in non-commutative (associative, unital) rings. An easy example is given by triangular rings (see for instance pp 17-22, especially p. 18 of Lam's First Course in Non-commutative Rings).

You can easily construct a finite counterexample with 8 elements: Let a be the ring of upper-triangular matrices; let i be the (two-sided) ideal of a consisting of strictly upper triangular matrices, and let j be the (two-sided) ideal consisting of matrices whose second row is zero. Then ij = 0, but ji = i. This works over any field, in particular the field with 2 elements.

For another example, consider the ring $\mathbb{Z}\langle x,y\rangle$ be the ring of polynomials with integer coefficients in two noncommuting variables $x$ and $y$ (this is the same as the integral semigroup ring of the free monoid of rank $2$). It's almost like the usual polynomial ring, except that the monomials are given by words in $x$ and $y$; so that, for example, the weight two monomials are $x^2$, $xy$, $yx$, and $y^2$, all distinct; the weight three monomials are $x^3$, $x^2y$, $xyx$, $yx^2$, $xy^2$, $yxy$, $y^2x$, and $y^3$, all distinct; etc.

Let $I = (x)$, the principal ideal generated by $x$. It consists of all elements of the form
$$\sum_{i=1}^n p_i(x,y)xq_i(x,y)$$
with $p_i,q_i\in\mathbb{Z}\langle x,y\rangle$, $n\geq 0$; i.e., all polynomials in which every monomial has an $x$ in it, somewhere. Likewise, $J=(y)$ consists of all polynomials in which every monomial has a $y$ in it somewhere.

Now, clearly, $xy\in IJ$. But it cannot be in $JI$, because every monomial in every nonzero element of $JI$ must have an $x$ that follows its first $y$, which is not the case for $xy$.